Question:
A proton accelerated by a potential difference $V$ gets
into the uniform electric field of a parallel-plate capacitor whose
plates extend over a length $l$ in the motion direction. The field
strength varies with time as $E = at$, where $a$ is a constant. Assuming
the proton to be non-relativistic, find the angle between the motion
directions of the proton before and after its flight through the
capacitor; the proton gets in the field at the moment $t = 0$. The edge
effects are to be neglected.

Answer: $\frac{al^{2}}{4}\sqrt{\frac{m}{2eV^{3}}}$

Solution:
The electric field inside the capacitor varies with time
as,
$$E = at$$
Hence, electric force on the proton,
$$F = eat$$
and subsequently, acceleration of the proton,
$$w = \frac{eat}{m}$$
Now, if $t$ is the time elapsed during the motion of the proton
between the plates, then $t = \frac{l}{v_{||}}$, as no acceleration is
effective in this direction. (Here $v_{||}$ is velocity along the
length of the plate.)
From kinematics, $\frac{dv_{\bot}}{dt} = w$
So, $\int_{0}^{t}{dv_{\bot}} = \int_{0}^{t}{wdt,}$
(as initially, the component of velocity in the direction, $\bot$ to
plates, was zero.)
Or
$v_{\bot} = \int_{0}^{t}\frac{ea}{m}\frac{t^{2}}{2m} = \frac{ea}{2m}\frac{l^{2}}{v_{||}^{2}}$
Now,
$\tan\alpha = \frac{v_{\bot}}{v_{||}} = \frac{eal^{2}}{2m\ v_{||}^{3}}$
$= \frac{eal^{2}}{2m\left( \frac{2eV}{m} \right)^{\frac{3}{2}}},$ as
$v_{||} = \left( \frac{2eV}{m} \right)^{\frac{1}{2}}$, from energy
conservation.
$$= \frac{al^{2}}{4}\sqrt{\frac{m}{2eV^{3}}}$$

Physics


Question:
A long solenoid of cross-sectional radius $a$ has a
thin insulated wire ring tightly put on its winding; one half of the
ring has the resistance $\eta$ times that of the other half. The
magnetic induction produced by the solenoid varies with time as
$B = bt$, where $b$ is a constant. Find the magnitude of the electric
field strength in the ring.

Answer: $\frac{1}{2}ab\frac{\eta - 1}{\eta + 1}$

Solution:
The changing magnetic field will induce an e.m.f. in the
ring, which is obviously equal, in the two parts by symmetry (the e.m.f.
induced by electromagnetic induction does not depend on resistance). The
current, that will flow due to this, will be different in the two parts.
This will cause an acceleration of charge, leading to the setting up of
an electric field $E$ which has opposite sign in the two parts. Thus,
$\frac{\xi}{2} - \pi aE = rl$ and, $\frac{\xi}{2} + \pi aE = \eta rI$
where $\xi$ is the total induced e.m.f. From this,
$\xi = (\eta + 1)rI$,
and
$E = \frac{1}{2\pi a}(\eta - 1)rI = \frac{1}{2\pi a}\frac{\eta - 1}{\eta + 1}\xi$
But by Faraday's law, $\xi = \pi a^{2}b$
So, $E = \frac{1}{2}ab\frac{\eta - 1}{\eta + 1}$

Physics


Question:
A long dielectric cylinder of radius $R$ is statically
polarized so that at all its points the polarization is equal to
$P = \alpha r$, where $\alpha$ is a positive constant, and $r$ is the
distance from the axis. The cylinder is set into rotation about its axis
with an angular velocity $\omega$. Find the magnetic induction $B$ at
the centre of the cylinder.

Answer: $B = B_{s} + B_{v} = 0$

Solution:
Because of polarization a space charge is present within
the cylinder. It's density is
$$\rho_{p} = - div\overrightarrow{P} = - 2\alpha$$
Since the cylinder as a whole is neutral a surface charge density
$\sigma_{P}$ must be present on the surface of the cylinder also. This
has the magnitude (algebraically)
$\sigma_{p} \times 2\pi R = 2\alpha\pi R^{2}$ or,
$\sigma_{p} = \alpha R$
When the cylinder rotates, currents are set up which give rise to
magnetic fields. The contribution of $\rho_{p}$ and $\sigma_{p}$ can
be calculated separately and then added.
For the surface charge the current is (for a particular element)
$$\alpha R \times 2\pi R\ dx \times \frac{\omega}{2\pi} = \alpha R^{2}\omega dx$$
Its contribution to the magnetic field at the centre is
$$\frac{\mu_{0}R^{2}\left( \alpha R^{2}\omega dx \right)}{2\left( x^{2} + R^{2} \right)^{3/2}}$$
and the total magnetic field is
$$B_{S} = \int_{\infty}^{- \infty}\frac{\mu_{0}R^{2}\left( \alpha R^{2}\omega dx \right)}{2\left( x^{2} + R^{2} \right)^{3/2}} = \frac{\mu_{0}\alpha R^{4}\omega}{2}\int_{- \infty}^{\infty}\frac{dx}{\left( x^{2} + R^{2} \right)^{3/2}} = \frac{\mu_{0}\alpha R^{4}\omega}{2} \times \frac{2}{R^{2}} = \mu_{0}\alpha R^{2}\omega$$
As for the volume charge density consider a circle of radius r, radial
thickness $dr$ and length $dx$.
The current is
$- 2\alpha \times 2\pi rdrdx \times \frac{\omega}{2\pi} = - 2\alpha r\ dr\ \omega dx$
The total magnetic field due to the volume charge distribution is
$$B_{v} = - \int_{0}^{R}{dr}\int_{- \infty}^{\infty}{dx}2\pi r\omega\frac{\mu_{0}r^{2}}{2\left( x^{2} + r^{2} \right)^{\frac{3}{2}}} = - \int_{0}^{R}{\alpha\mu_{0}\omega r^{3}\ dr}\int_{- \infty}^{\infty}{dx}\left( x^{2} + r^{2} \right)^{3/2}$$
$$= - \int_{0}^{R}{\alpha\mu_{0}\omega r}dr \times 2 = - \mu_{0}\alpha\omega R^{2}\ \ so,B = B_{s} + B_{v} = 0$$

Physics


Question:
Inside an infinitely long circular cylinder charged
uniformly with volume density $\rho$ there is a circular cylindrical
cavity. The distance between the axes of the cylindrical cavity. The
distance between the axes of the cylinder and the cavity is equal to a.
Find the electric field strength E inside the cavity. The permittivity
is assumed to be equal to unity.

Answer: $\overrightarrow{E} = \overrightarrow{E} + \overrightarrow{E} = \frac{\rho}{2\varepsilon_{0}}\left( \overrightarrow{r_{+}} - \overrightarrow{r_{-}} \right) = \frac{\rho}{2\varepsilon_{0}}\overrightarrow{a}$

Solution:
Let us consider a cylindrical Gaussian surface of radius
$r$ and height $h$ inside an infinitely long charged cylinder with
charge density $\rho$. Now from Gauss theorem:
$$E_{r}2\pi rh = \frac{q_{inclosed}}{\varepsilon_{0}}$$
(where $E_{r}$ is the field inside the cylinder at a distance $r$ from
it axis.)
Or, $E_{r}2\pi rh = \frac{\rho\pi r^{2}h}{\varepsilon_{0}}$ or
$E_{r} = \frac{\rho r}{2\varepsilon_{0}}$
Now, using the method of 3.28 field at a point $P$, inside the cavity,
is
$$\overrightarrow{E} = \overrightarrow{E} + \overrightarrow{E} = \frac{\rho}{2\varepsilon_{0}}\left( \overrightarrow{r_{+}} - \overrightarrow{r_{-}} \right) = \frac{\rho}{2\varepsilon_{0}}\overrightarrow{a}$$
[IMAGE]

Physics


Question:
Find the pressure of air in a vessel being evacuated as
a function of evacuation time $t$. The vessel volume is $V$, the initial
pressure is $p_{0}$. The process is assumed to be isothermal, and the
evacuation rate equal to $C$ and independent of pressure.
Note: The evacuations rate is the gas volume being evacuated per
unit time, with that volume being measured under the gas pressure
attained by that moment.

Answer: $p = p_{0}e^{- Ct/V}$

Solution:
From the ideal gas equation $p = \frac{m}{M}\frac{RT}{V}$
$\frac{dp}{dt} = \frac{RT}{MV}\frac{dm}{dt}$ (1)
In each stroke, volume $v$ of the gas is ejected, where $v$ is given
by
$$v = \frac{V}{m_{N}}\left\lbrack m_{N - 1} - m_{N} \right\rbrack$$
In case of continuous ejection, if $\left( m_{N - 1} \right)$
corresponds to mass of gas in the vessel at time $t$, time $m_{N}$ is
the mas at time $t + \mathrm{\Delta}t$, where $\mathrm{\Delta}t$, is
the time in which volume $v$ of the gas has come out. The rate of
evacuation is therefore $\frac{v}{\mathrm{\Delta}t}$ i.e.
$$C = \frac{v}{\mathrm{\Delta}t} = - \frac{V}{m(t + \mathrm{\Delta}t)}.\frac{m(t + \mathrm{\Delta}t) - m(t)}{\mathrm{\Delta}t}$$
In the limit $\mathrm{\Delta}t - 0$, we get
$C = \frac{V}{m}\frac{dm}{dt}$ (2)
From (1) and (2)
$$\frac{dp}{dt} = - \frac{C}{V}\frac{mRT}{MV} = - \frac{C}{V}p\ \ \ \ or\ \ \ \frac{dp}{p} = - \frac{C}{V}dt$$
Integrating
$\int_{p}^{p_{0}}\frac{dp}{p} = - \frac{C}{V}\int_{t}^{0}{dt}$ or
$\ln\frac{p}{p_{0}} = - \frac{C}{V}t$
Thus $p = p_{0}e^{- Ct/V}$

Physics


Question:
A cylinder rolls without slipping over a horizontal
plane. The radius of the cylinder is equal to $r$. Find the curvature
radii of trajectories traced out by the points $A$ and $B$ (see Fig.).
[IMAGE]

Answer: $2\sqrt{2}r$

Solution:
Let us draw the kinematical diagram of the rolling
cylinder on the basis of the solution of problem 1.53.
[IMAGE] Figure 1.54: Insert this diagram from the shared reference image ---
rolling cylinder kinematics and acceleration directions.
As, an arbitrary point of the cylinder follows a curve, its normal
acceleration and radius of curvature are related by the well known
equation
$$w_{n} = \frac{v^{2}}{R}$$
so, for point $A$,
$$w_{A(n)} = \frac{v_{A}^{2}}{R_{A}}$$
or,
$$R_{A} = \frac{4v_{c}^{2}}{\omega^{2}r} = 4r$$
because $v_{c} = \omega r$, for pure rolling.
Similarly for point $B$,
$$w_{B(n)} = \frac{v_{B}^{2}}{R_{B}}$$
$$\omega^{2}r\cos 45^{\circ} = \frac{\left( \sqrt{2}v_{c} \right)^{2}}{R_{B}},$$
or,
$$R_{B} = 2\sqrt{2}\frac{v_{c}^{2}}{\omega^{2}r} = 2\sqrt{2}r$$

Physics


Question:
A system consists of a long cylindrical anode of radius
$a$ and a coaxial cylindrical cathode of radius $b(b < a)$. A filament
located along the axis of the system carries a heating current $I$
producing a magnetic field in the surrounding space. Find the least
potential difference between the cathode and anode at which the thermal
electrons leaving the cathode without initial velocity start reaching
the anode.

Answer: $V \geq \frac{1}{2}\frac{|e|}{m}\left( \frac{\mu_{0}I}{2\pi} \right)^{2}\ln\frac{a}{b}$

Solution:
When a current $I$ flows along the axis, a magnetic field
$B_{\varphi} = \frac{\mu_{0}I}{2\pi\rho}$ is set up where
$\rho^{2} = x^{2} + y^{2}$. In terms of components,
$B_{x} = - \frac{\mu_{0}Iy}{2\pi\rho^{2}},B_{y} = \frac{\mu_{0}Ix}{2\pi\rho^{2}}$
and $B_{z} = 0$
Suppose a p.d. $V$ is set up between the inner cathode and the outer
anode. This means a potential function of the form
$$\varphi = V\frac{\ln{\rho/b}}{\ln{a/b}},a > \rho > b,$$
as one can check by solving Laplace equation.
[IMAGE] The electric field corresponding to this is,
$$E_{x} = - \frac{Vx}{\rho^{2}\ln{a/b}},E_{y} = - \frac{Vy}{\rho^{2}\ln{a/b}},E_{z} = 0$$
The equations of motion are,
$$\frac{d}{dt}mv_{x} = + \frac{|e|Vz}{\rho^{2}\ln{a/b}} + \frac{|e|\mu_{0}I}{2\pi\rho^{2}}x\acute{z}$$
$$\frac{d}{dt}mv_{y} = + \frac{|e|Vy}{\rho^{2}\ln{a/b}} + \frac{|e|\mu_{0}I}{2\pi\rho^{2}}y\acute{z}$$
and
$\frac{d}{dt}mv_{z} = - |e|\frac{\mu_{0}I}{2\pi\rho^{2}}\left( x\acute{x} + y\acute{y} \right) = - |e|\frac{\mu_{0}I}{2\pi}\frac{d}{dt}\ln\rho$
$\left( - |e| \right)$ is the charge on the electron.
Integrating the last equation,
$$mv_{z} = - |e|\frac{\mu_{0}I}{2\pi}\ln{\rho/a} = m\acute{z}$$
since $v_{z} = 0$ where $\rho = a$. We now substitute this $\acute{z}$
in the other two equations to get
$$\frac{d}{dt}\left( \frac{1}{2}mv_{x}^{2} + \frac{1}{2}mv_{y}^{2} \right)$$
$$= \left\lbrack \frac{|e|V}{\ln\frac{a}{b}} - \frac{|e|^{2}}{m}\left( \frac{\mu_{0}I}{2\pi} \right)^{2}\ln\frac{\rho}{b} \right\rbrack.\frac{x\acute{x} + y\acute{y}}{\rho^{2}}$$
$$= \left\lbrack \frac{|e|V}{\ln\frac{a}{b}} - \frac{|e|^{2}}{m}\left( \frac{\mu_{0}I}{2\pi} \right)^{2}\ln\frac{\rho}{b} \right\rbrack \bullet \frac{1}{2\rho^{2}}\frac{d}{dt}\rho^{2}$$
$$= \left\lbrack \frac{|e|V}{\ln\frac{a}{b}} - \frac{|e|^{2}}{m}\left( \frac{\mu_{0}I}{2\pi} \right)^{2}\ln\frac{\rho}{b} \right\rbrack\frac{d}{dt}\ln\frac{\rho}{b}$$
Integrating and using $v^{2} = 0,$ at $\rho = b,$ we get,
$$\frac{1}{2}mv^{2} = \left\lbrack \frac{|e|V}{\ln\frac{a}{b}}\ln\frac{\rho}{b} - \frac{1}{2m}|e|^{2}\left( \frac{\mu_{0}I}{2\pi} \right)^{2}\left( \ln\frac{\rho}{b} \right) \right\rbrack$$
The RHS must be positive, for all $a > \rho > b$. The condition for
this is,
$$V \geq \frac{1}{2}\frac{|e|}{m}\left( \frac{\mu_{0}I}{2\pi} \right)^{2}\ln\frac{a}{b}$$

Physics


Question:
A standing wave $y = a\sin{kx}.\cos{\omega t}$ is
maintained in a homogeneous rod with cross-sectional area $S$ and
density $\rho$. Find the total mechanical energy confined between the
sections corresponding to the adjacent displacement nodes.

Answer: $E = T + U = \frac{\pi\rho\omega^{2}a^{2}S}{4k}$

Solution:
We have a standing wave given by the equation
$$y = a\sin{kx}\cos{\omega t}$$
So, $\frac{\partial\xi}{\partial t} = - a\omega\sin kx\sin\omega t$
(1)
and $\frac{\partial y}{\partial t} = ak\cos kx\cos\omega t$ (2)
The kinetic energy confined in an element of length $dx$ of the rod
$$dT = \frac{1}{2}(\rho Sdx)\left( \frac{\partial y}{\partial t} \right)^{2} = \frac{1}{2}\rho Sa^{2}\omega^{2}\sin^{2}kx\ dx$$
So total kinetic energy confined into rod
$$T = \int_{}^{}{dT} = \frac{1}{2}\rho Sa^{2}\omega^{2}\sin^{2}{\omega t}\int_{0}^{\lambda/2}{\sin^{2}\frac{2\pi}{\lambda}x}dx$$
Or, $T = \frac{\pi Sa^{2}\omega^{2}\rho\sin^{2}{\omega t}}{4k}$ (3)
The potential energy in the above rod element
$dU = \int_{}^{}{\partial U} = - \int_{0}^{y}{F_{y}dy}$, where
$F_{y} = (\rho Sdx)\frac{\partial^{2}y}{\partial t^{2}}$
Or, $F_{y} = - (\rho Sdx)\omega^{2}y$
So, $dU = \omega^{2}\rho Sdx\int_{0}^{y}{ydy}$
Or,
$dU = \frac{\rho\omega^{2}Sy^{2}}{2}dx = \frac{\rho\omega^{2}Sa^{2}\cos^{2}\ \omega t\sin^{2}{kx}dx}{2}$
Thus the total potential energy stored in the rod $U = \int_{}^{}{dU}$
Or,
$U = \rho\omega^{2}Sa^{2}\cos^{2}{\omega t}\int_{0}^{\lambda/2}{\sin^{2}{\frac{2\pi}{\lambda}x}}dx$
So, $U = \frac{\pi\rho Sa^{2}\omega^{2}\cos^{2}{\omega t}}{4k}$
To find the potential energy stored in the rod element we may adopt an
easier way. We know that the potential energy density confined in a
rod under elastic force equals:
$$U_{D} = \frac{1}{2}(stress \times strain) = \frac{1}{2}\sigma\varepsilon = \frac{1}{2}Y\varepsilon^{2}$$
$$= \frac{1}{2}\rho v^{2}\varepsilon^{2} = \frac{1}{2}\frac{\rho\omega^{2}}{k^{2}}\varepsilon^{2}$$
$$= \frac{1}{2}\frac{\rho\omega^{2}}{k^{2}}\left( \frac{\partial y}{\partial x} \right)^{2} = \frac{1}{2}\rho a^{2}\omega^{2}\cos^{2}{\omega t}\cos^{2}{kx}$$
Hence the total potential energy stored in the rod
$$U = \int_{}^{}U_{D}dV = \int_{0}^{\lambda/2}\frac{1}{2}\rho a^{2}\omega^{2}\cos^{2}{\omega t}\cos^{2}{kx}Sdx$$
$$= \frac{\pi\rho Sa^{2}\omega^{2}\cos^{2}{\omega t}}{4k}$$
Hence the sought mechanical energy confined in the rod between the two
adjacent nodes
$$E = T + U = \frac{\pi\rho\omega^{2}a^{2}S}{4k}$$

Physics


Question:
A parallel-plate capacity whose electrodes are shaped
as round discs is charged slowly. Demonstrate that the flux of the
Poynting vector across the capacitor's lateral surface is equal to the
increment of the capacitor's energy per unit time. The dissipation of
field at the edge is to be neglected in calculations.

Answer: $2\pi Rd \times S = \frac{Q\dot{Q}d}{\pi R^{2}\varepsilon_{0}}$

Solution:
If the charge on the capacitor is $Q$, the rate of
increase of the capacitor's energy
$$= \frac{d}{dt}\left( \frac{1}{2}\frac{Q^{2}}{C} \right) = \frac{Q\dot{Q}}{C} = \frac{d}{\varepsilon_{0}\pi R^{2}}Q\dot{Q}$$
Now electric field between the plates (inside it) is,
$E = \frac{Q}{\pi R^{2}\varepsilon_{0}}$.
So displacement current
$= \frac{\partial D}{\partial t} = \frac{\dot{Q}}{\pi R^{2}}$
This will lead to a magnetic field, (circuital) inside the plates. At
a radial distance $r$
$2\pi rH_{\theta}(r) = \pi r^{2}\frac{\dot{Q}}{\pi R^{2}}$ or
$H_{\theta} = \frac{\dot{Q}r}{2\pi R^{2}}$
Hence $H_{\theta}(R) = \frac{Q}{2\pi R}$ at the edge.
Thus inward Poynting vector
$= S = \frac{Q}{2\pi R} \times \frac{Q}{\pi R^{2}\varepsilon_{0}}$
Total flow
$= 2\pi Rd \times S = \frac{Q\dot{Q}d}{\pi R^{2}\varepsilon_{0}}$
Proved

Physics


Question:
An inhomogeneous poorly conducting medium fills up the
space between plates 1 and 2 of a parallel-plate capacitor. Its
permittivity and resistivity vary from values $\varepsilon_{1},\rho_{1}$
at plate 1 to values $\varepsilon_{2},\rho_{2}$ at plate 2. A dc voltage
is applied to the capacitor through which a steady current I flows from
plate 1 to plate 2. Find the total extraneous charge in the given
medium.

Answer: $\varepsilon_{0}I\left( \varepsilon_{2}\rho_{2} - \varepsilon_{1}\rho_{1} \right)$

Solution:
By current conservation
$$\frac{E(x)}{\rho(x)} = \frac{E(x) + dE(x)}{\rho(x) + d\rho(x)} = \frac{dE(x)}{d\rho(x)}$$
This has the solution
$$E(x) = C\rho(x) = \frac{I\rho(x)}{A}$$
[IMAGE] Hence charge induced in the slice per unit area
$$d\sigma = \varepsilon_{0}\frac{I}{A}\left\lbrack \left\{ \varepsilon(x) + d\varepsilon(x) \right\}\left\{ \rho(x) + d\rho(x) \right\} - \varepsilon(x)\rho(x) \right\rbrack = \varepsilon_{0}\frac{I}{A}d\left\lbrack \varepsilon(x)\rho(x) \right\rbrack$$
Thus,
$dQ = \varepsilon_{0}Id\left\lbrack \varepsilon(x)\rho(x) \right\rbrack$
Hence total charge induced, is by integration,
$$Q = \varepsilon_{0}I\left( \varepsilon_{2}\rho_{2} - \varepsilon_{1}\rho_{1} \right)$$

Physics


Question:
A man standing on the edge of a swimming pool looks at
a stone lying on the bottom. The depth of the swimming pool is equal to
$h$. At what distance from the surface of water is the image of the
stone formed if the line of vision makes an angle $\theta$ with the
normal to the surface?

Answer: $\frac{n^{2}h\cos^{3}\theta}{\left( n^{2} - \sin^{2}\theta \right)^{3/2}}$

Solution:
From the Fig.
[IMAGE] $$\sin{d\alpha} = \frac{MP}{OM} = \frac{MN\cos\alpha}{h\sec(\alpha + d\alpha)}$$
As $d\alpha$ is very small, so
$d\alpha = \frac{MN\cos\alpha}{h\sec\alpha} = \frac{MN\cos^{2}\alpha}{h}$
(1)
Similarly
$d\theta = \frac{MN\cos^{2}\theta}{h'}$ (2)
From Eqns (1) and (2)
$\frac{d\alpha}{d\theta} = \frac{h^{'}\cos^{2}\alpha}{h\cos^{2}\theta}$
or,
$h^{'} = \frac{h\cos^{2}\theta}{\cos^{2}\alpha}\frac{d\alpha}{d\theta}$
(3)
From the law of refraction
$$n\sin\alpha = \sin\theta$$
$\sin\alpha = \frac{\sin\theta}{n},$ so,
$\cos\alpha = \sqrt{\frac{n^{2} - \sin^{2}\theta}{n^{2}}}$
Differentiating Eqn. (A)
$n\cos\alpha d\alpha = \cos\theta d\theta$ or,
$\frac{d\alpha}{d\theta} = \frac{\cos\theta}{n\cos\alpha}$ (4)
Using (4) and (3), we get
$h^{'} = \frac{h\cos^{3}\theta}{n\cos^{3}\alpha}$ (5)
Hence
$h^{'} = \frac{h\cos^{3}\theta}{n\left( \frac{n^{2} - \sin^{2}\theta}{n^{2}} \right)^{3/2}} = \frac{n^{2}h\cos^{3}\theta}{\left( n^{2} - \sin^{2}\theta \right)^{3/2}}$
\[Using Eqn. (B)\]

Physics


Question:
A metal rod of mass $m$ can rotate abut a horizontal
axis $O$, sliding along a circular conductor of radius. The arrangement
is located in a uniform magnetic field of induction $B$ directed
perpendicular to the ring plane. The axis and the ring are connected to
an emf source to form a circuit of resistance $R$. Neglecting the
friction, circuit inductance, and ring resistance, find the law
according to which the source emf must vary to make the rod rotate with
a constant angular velocity $\omega$.
[IMAGE]

Answer: $\frac{1}{2aB}\left( a^{3}B^{2}\omega + 2mgR\sin\omega t \right)$

Solution:
As the rod rotates, an emf.
$$\frac{d}{dt}\frac{1}{2}a^{2}\theta \bullet B = \frac{1}{2}a^{2}B\omega$$
is induced in it. The net current in the conductor is then
$\frac{\varepsilon(t) - \frac{1}{2}a^{2}B\omega}{R}$
A magnetic force will then act on the conductor of magnitude BI per
unit length. Its direction will be normal to B and rod and its torque
will be
$$\int_{0}^{a}{\left( \frac{\varepsilon(t) - \frac{1}{2}a^{2}B\omega}{R} \right)dx}Bx$$
Obviously both magnetic and mechanical torque acting on the C.M. of
the rod must be equal but opposite in sense. Then
for equilibrium at constant $\omega$
$$\frac{\varepsilon(t) - \frac{1}{2}a^{2}B\omega}{R} \bullet \frac{Ba^{2}}{2} = \frac{1}{2}mga\sin{\omega t}$$
Or,
$\varepsilon(t) = \frac{1}{2}a^{2}B\omega + \frac{mgR}{aB}\sin{\omega t} = \frac{1}{2aB}\left( a^{3}B^{2}\omega + 2mgR\sin\omega t \right)$
(The answer given in the book is incorrect dimensionally.)

Physics


Question:
A parallel-plate capacitor was lowered into water in a
horizontal position, with water filling up the gap between the plates
$d = 1.0\ mm$ wide. Then a constant voltage $V = 500\ V$ was applied to
the capacitor. Find the water pressure increment in the gap.

Answer: $\mathrm{\Delta}p = 7.17\ kPa = 0.07\ atm$

Solution:
When the capacitor which is immersed in water is connected
to a constant voltage source, it gets charged. Suppose $\sigma_{0}$ is
the free charge density on the condenser plates. Because water is a
dielectric, bound charges also appear in it. Let $\sigma'$ be the
surface density of bound charges. (Because of homogeneity of the medium
and uniformity of the field when we ignore edge effects no volume
density of bound charges exists.) The electric field due to free charges
only $\frac{\sigma_{0}}{\varepsilon_{0}}$; that due to bound charges is
$\frac{\sigma'}{\varepsilon_{0}}$ and the total electric field a
$\frac{\sigma_{0}}{\varepsilon\varepsilon_{0}}$. Recalling that the sign
of bound charges is opposite of the free charges, we have
$\frac{\sigma_{0}}{\varepsilon\varepsilon_{0}} = \frac{\sigma_{0}}{\varepsilon_{0}} - \frac{\sigma^{'}}{\varepsilon_{0}}$
or,
$\sigma^{'} = \left( \frac{\varepsilon - 1}{\varepsilon} \right)\sigma_{0}$
Because of the field that exists due to the free charges (not the
total field; the field due to the bound charges must be excluded for
this purpose as they only give rise to self energy effects), there is
a force attracting the bound charges to the near by plates. This force
is
$\frac{1}{2}\sigma^{'}\frac{\sigma_{0}}{\varepsilon_{0}} = \frac{(\varepsilon - 1)\sigma_{0}^{2}}{2\varepsilon\varepsilon_{0}}$
per unit area.
The factor $\frac{1}{2}$ needs an explanation. Normally the force on a
test charge is qE in an electric field $E$. But if the charge itself
is produced by the electric filed then the force must be constructed
bit by bit and is
$$F = \int_{0}^{E}{q\left( E^{'} \right)}dE'$$
if $q(E') \propto E'$ then we get
$$F = \frac{1}{2}q(E)E$$
This factor of $\frac{1}{2}$ is well known. For example the energy of
a dipole of moment $\overrightarrow{p}$ in an electric field
${\overrightarrow{E}}_{0}$ is
$- \overrightarrow{p} \bullet {\overrightarrow{E}}_{0}$ while the
energy per unit volume of a linear dielectric in an electric field is
$- \frac{1}{2}\overrightarrow{P} \bullet {\overrightarrow{E}}_{0}$
where $\overrightarrow{P}$ is the polarization vector (i.e. dipole
moment per unit volume). Now the force per unit area manifests itself
as excess pressure of the liquid.
Noting that
$\frac{V}{d} = \frac{\sigma_{0}}{\varepsilon\varepsilon_{0}}$
We get
$\mathrm{\Delta}p = \frac{\varepsilon_{0}\varepsilon(\varepsilon - 1)V^{2}}{2d^{2}}$
Substitution, using $\varepsilon = 81$ for water, gives
$\mathrm{\Delta}p = 7.17\ kPa = 0.07\ atm$.

Physics


Question:
A fluid with viscosity $\eta$ fills the space between
two long co-axial cylinders of radii $R_{1}$ and $R_{2}$, with
$R_{1} < R_{2}$. The inner cylinder is stationary while the outer one is
rotated with a constant angular velocity $\omega_{2}$. The fluid flow is
laminar. Taking into account that the friction force acting on a unit
area of a cylindrical surface of radius $r$ is defined by the formula
$\sigma = \eta r(\partial\omega/\partial r)$, find :
\(a\) the angular velocity of the rotating fluid as a function of radius
$r$;
\(b\) the moment of the friction forces acting on a unit length of the
outer cylinder.

Answer:
(a)
$\omega = \omega_{2}\frac{R_{1}^{2}R_{2}^{2}}{R_{2}^{2} - R_{1}^{2}}\left\lbrack \frac{1}{R_{1}^{2}} - \frac{1}{r^{2}} \right\rbrack$
\(b\)
$N_{1} = \frac{N}{l} = 4\pi\eta\omega_{2}\frac{R_{1}^{2}R_{2}^{2}}{R_{2}^{2} - R_{1}^{2}}$

Solution:
(a) Let us consider an elemental cylinder of radius $r$
and thickness $dr$ then from Newton's formula
$$F = 2\pi rl\eta r\frac{d\omega}{dr} = 2\pi l\eta r^{2}\frac{d\omega}{dr}$$
and moment of this force acting on the element,
$$N = 2\pi r^{2}l\eta\frac{d\omega}{dr}r = 2\pi r^{3}l\eta\frac{d\omega}{dr}$$
Or, $2\pi l\eta d\omega = N\frac{dr}{r^{3}}$ (2)
[IMAGE] As in the previous problem $N$ is constant when conditions are steady
Integrating,
$2\pi l\eta\int_{0}^{\omega}{d\omega} = N\int_{R_{1}}^{r}\frac{dr}{r^{3}}$
Or,
$2\pi l\eta\omega = \frac{N}{2}\left\lbrack \frac{1}{R_{1}^{2}} - \frac{1}{r^{2}} \right\rbrack$
(3)
Putting $r = R_{2},\omega = \omega_{2}$, we get
$2\pi l\eta\omega_{2} = \frac{N}{2}\left\lbrack \frac{1}{R_{1}^{2}} - \frac{1}{R_{2}^{2}} \right\rbrack$
(4)
From (3) and (4),
$$\omega = \omega_{2}\frac{R_{1}^{2}R_{2}^{2}}{R_{2}^{2} - R_{1}^{2}}\left\lbrack \frac{1}{R_{1}^{2}} - \frac{1}{r^{2}} \right\rbrack$$
\(b\) From Eq. (4),
$N_{1} = \frac{N}{l} = 4\pi\eta\omega_{2}\frac{R_{1}^{2}R_{2}^{2}}{R_{2}^{2} - R_{1}^{2}}$

Physics


Question:
At what minimum kinetic energy must a hydrogen atom
move for its inelastic head on collision with another, stationary,
hydrogen atom to make one of them capable of emitting a photon ? Both
atoms are supposed to be in the ground state prior to the collision.

Answer: $20.4\ eV$

Solution:
Photon can be emitted in $H - H$ collision only if one of
the $H$ is excited to an $n = 2$ state which then dexcites to $n = 1$
state by emitting a photon. Let $v_{1}$ and $v_{2}$ be the velocities of
the two Hydrogen atoms after the collision and $M$ their masses. Then,
energy momentum conservation
$$Mv_{1} + Mv_{2} = \sqrt{2MT}$$
(in the frame of the stationary $H$ atom)
$$\frac{1}{2}Mv_{1}^{2} + \frac{1}{2}Mv_{2}^{2} + \frac{3}{4}hR = T$$
$\frac{3}{4}hR = hR\left( 1 - \frac{1}{4} \right)$ is the excitation
energy of the $n = 2$ state from the ground state.
Eliminating $v_{2}$
$\frac{1}{2}M\left\{ v_{1}^{2} + \left( \sqrt{\frac{2T}{M}} - v_{1} \right)^{2} \right\} + \frac{3}{4}hR = T$
Or
$\frac{1}{2}M\left\{ 2v_{1}^{2} - 2\sqrt{\frac{2T}{M}}v_{1} + \frac{2T}{M} \right\} + \frac{3}{4}hR = T$
$M\left\{ \left( v_{1} - \frac{1}{2}\sqrt{\frac{2T}{M}} \right)^{2} \right\} + \frac{1}{2}T + \frac{3}{4}hR = T$
Or
$M\left\{ \left( v_{1} - \frac{1}{2}\sqrt{\frac{2T}{M}} \right) \right\} + \frac{3}{4}hR = \frac{1}{2}T$
For minimum $T$, the square on the left should vanish. Thus
$T = \frac{3}{2}hR = 20.4\ eV$

Physics


Question:
An alpha particle with kinetic energy $T = 0.50\ MeV$
is deflected through an angle of $\theta = 90{^\circ}$ by the Coulomb
field of a stationary Hg nucleus. Find:
All the formulas in this Part are given the Gaussian system of units.
\(a\) the least curvature radius of its trajectory;
\(b\) the minimum approach distance between the particle and the
nucleus.

Answer:
(a) $0.231\ pm$
\(b\) $0.557\ pm$

Solution:
We shall ignore the recoil of Hg nucleus.
\(a\) Let $A$ be the point of closest approach to the centre
$C,AC = r_{\min}$. At $A$ the motion is instantaneously circular
because the radial velocity vanishes. Then if $v_{0}$ is the speed of
the particle at $A,$ the following equations hold
$\Gamma = \frac{1}{2}mv_{0}^{2} + \frac{Z_{1}Z_{2}e^{2}}{\left( 4\pi\varepsilon_{0} \right)r_{\min}}$
(1)
$mv_{0}r_{\min} = \sqrt{2mT}b$ (2)
$\frac{mv_{0}^{2}}{\rho_{\min}} = \frac{Z_{1}Z_{2}e^{2}}{\left( 4\pi\varepsilon_{0} \right)r_{\min}^{2}}$
(3)
[IMAGE] (This is Newton's law. Here $\rho = \rho_{\min}$ is the radius of
curvature of the path at A and $\rho$ is minimum at $A$ by symmetry.)
Finally we have Eqn. (6.1 a) in the form
$b = \frac{Z_{1}\ Z_{2}\ e^{2}}{\left( 4\pi\varepsilon_{0} \right)2T}\cot\frac{\theta}{2}$
(4)
From (2) and (3)
$\frac{2Tb^{2}}{\rho_{\min}} = \frac{Z_{1}\ Z_{2}\ e^{2}}{\left( 4\pi\varepsilon_{0} \right)}$
Or
$\rho_{\min} = \frac{Z_{1}\ Z^{2}\ e^{2}}{\left( 4\pi\varepsilon_{0} \right)2T}\cot^{2}\frac{\theta}{2},$
with $z_{1} = 2,z_{2} = 80$ we get
$\rho_{\min} = 0.231\ pm$
\(b\) From (2) and (4) we write
$r_{\min} = \frac{z_{1}\ z_{2}\ e^{2}}{\left( 4\pi\varepsilon_{0} \right)\sqrt{2m\ T}}\frac{\cot{\theta/2}}{v_{0}},$
Substituting in (1)
$T = \frac{1}{2}mv_{0}^{2} + \sqrt{2\ m\ T}v_{0}\tan{\theta/2}$
Solving for $v_{0}$ we get
$v_{0} = \sqrt{\frac{2T}{m}}\left( \sec\frac{\theta}{2} - \tan\frac{\theta}{2} \right)$
Then
$r_{\min} = \frac{z_{1}\ z_{2}\ e^{2}}{\left( 4\pi\varepsilon_{0} \right)\ 2T}\frac{\cot\frac{\theta}{2}}{\sec\frac{\theta}{2} - \tan\frac{\theta}{2}}$
$= \frac{z_{1}\ z_{2}\ e^{2}}{\left( 4\pi\varepsilon_{0} \right)2T}\cot\frac{\theta}{2}\left( \sec\frac{\theta}{2} + \tan\frac{\theta}{2} \right)$
$= \frac{z_{1}\ z_{2}\ e^{2}}{\left( 4\pi\varepsilon_{0} \right)2T}\left( 1 + {cosec}\frac{\theta}{2} \right) = 0.557\ pm$

Physics


Question:
Suppose the surface charge density over a sphere of
radius $R$ depends on a polar angle $\theta$ as
$\sigma = \sigma_{0} = \cos\theta$, where $\sigma_{0}$ is a positive
constant. Show that such a charge distribution can be represented as a
result of a small relative shift of two uniformly charged balls of
radius $R$ whose charges are equal in magnitude and opposite in sign.
Resorting to this representation, find the electric field strength
vector inside the given sphere.
$$= - \frac{\rho\overrightarrow{a}}{3\varepsilon_{0}} = \frac{\sigma_{0}}{3\varepsilon_{0}}\overrightarrow{k}$$

Answer: $\overrightarrow{E} = - \frac{\rho\overrightarrow{a}}{3\varepsilon_{0}} = \frac{\sigma_{0}}{3\varepsilon_{0}}\overrightarrow{k}$

Solution:
We start from two charged spherical balls each of radius
$R$ with equal and opposite charge densities $+ \rho$ and $- \rho$. The
centre of the balls are at $+ \frac{\overrightarrow{a}}{2}$ and
$- \frac{\overrightarrow{a}}{2}$ respectively so the equation of their
surface are
$\left| \overrightarrow{r} - \frac{\overrightarrow{a}}{2} \right| = R$
or $r - \frac{a}{2}\cos{\theta \simeq R}$ and
$r + \frac{a}{2}\cos\theta \simeq R$, considering a to be small. The
distance between the two surfaces in the radial direction at angle
$\theta$ is $\left| a\cos\theta \right|$ and does not depend on the
azimuthal angle. It is seen from the diagram that the surface of the
sphere has in effect a surface density $\sigma = \sigma_{0}\cos\theta$
when
$$\sigma_{0} = \rho a$$
Inside any uniformly charged spherical ball, the field is radial and
has the magnitude given by Gauss's theorm
$$4\pi r^{2}E = \frac{4\pi}{3}r^{3}\rho/\varepsilon_{0}$$
Or $E = \frac{\rho r}{3\varepsilon_{0}}$
[IMAGE] In vector notation, using the fact the $V$ must be measured from the
centre of the ball, we get, for the present case
$$\overrightarrow{E} = \frac{\rho}{3\varepsilon_{0}}\left( \overrightarrow{r} - \frac{\overrightarrow{a}}{2} \right) - \frac{\rho}{3\varepsilon_{0}}\left( \overrightarrow{r} + \frac{\overrightarrow{a}}{2} \right)$$
$$\overrightarrow{E} = - \frac{\rho\overrightarrow{a}}{3\varepsilon_{0}} = \frac{\sigma_{0}}{3\varepsilon_{0}}\overrightarrow{k}$$
When $\overrightarrow{k}$ is the unit vector along the polar axis from
which $\theta$ is measured.

Physics


Question:
A thin wire ring of radius $a$ and resistance $r$ is
located inside a long solenoid so that their axes coincide. The length
of the solenoid is equal to $l$, its cross-sectional radius, to $b$. At
a certain moment the solenoid was connected to a source of constant
voltage $V$. The total resistance of the circuit is equal to $R$.
Assuming the inductance of the ring to be negligible, find the maximum
value of the radial force acting per unit length of the ring.

Answer: $\frac{\mu_{0}a^{2}V^{2}}{4r\ Rlb^{2}}$

Solution:
The solenoid has an inductance,
$$L = \mu_{0}n^{2}\pi b^{2}I,$$
where $n =$ number of turns of the solenoid per unit length. When the
solenoid is connected to the source an e.m.f. is set up, which,
because of the inductance and resistance, rises slowly, according to
the equation,
$$RI + LI = V$$
This has the well known solution,
$$I = \frac{V}{R}\left( 1 - e^{- t\ R/L} \right)$$
Corresponding to this current, an e.m.f. is induced in the ring. Its
magnetic field $B = \mu_{0}nI$ in the solenoid, produces a force per
unit length, $\frac{dF}{dl} = Bi = \mu_{0}^{2}n^{2}\pi a^{2}\ II/r$
$$= \frac{\mu_{0}^{2}\pi a^{2}V^{2}}{r}\left( \frac{n^{2}}{RL} \right)e^{- t\ R/L}\left( 1 - e^{- tR/L} \right),$$
acting on each segment of the ring. This force is zero initially and
zero for large $t$. Its maximum value is for some finite $t$. The
maximum value of
$e^{- tR/L}\left( 1 - e^{- tR/L} \right) = \frac{1}{4} - \left( \frac{1}{2} - e^{- tR/L} \right)^{2}$
is $\frac{1}{4}$.
So
$\frac{dF_{\max}}{dl} = \frac{\mu_{0}^{2}\pi a^{2}V^{2}}{r}\frac{n^{2}}{4RL} = \frac{\mu_{0}a^{2}V^{2}}{4r\ Rlb^{2}}$

Physics


Question:
A plane light wave with wavelength
$\lambda = 0.70\ \mu m$ fails normally on the base of biprism made of
glass $(n = 1.520)$ with refracting angle $\theta = 5.0{^\circ}$. Behind
the biprism Fig. there is a plane-parallel plate, with the space between
them filled up with benzene $\left( n^{'} = 1.500 \right)$. Find the
width of a fringe on the screen $Sc$ placed behind this system.
[IMAGE]

Answer: $0.2\ mm$

Solution:
It will be assumed that the space between the biprism and
the glass plate filled with benzene constitutes complementary prisms as
shown.
[IMAGE] Then the two prisms being oppositely placed, the net deviation
produced by them is
$$\delta = (n - 1)\theta - \left( n^{'} - 1 \right)\theta = \left( n - n^{'} \right)\theta$$
Hence as in the previous problem
$$d = 2a\delta = 2a\theta\left( n - n^{'} \right)$$
So,
$\mathrm{\Delta}x = \frac{(a + b)\lambda}{2a\theta\left( n - n^{'} \right)}$
For plane incident wave we let $a \rightarrow \infty$
So, $\mathrm{\Delta}x = \frac{\lambda}{2\theta(n - n')} = 0.2\ mm$

Physics


Question:
Find the velocity of the excited hydrogen atoms if the
first line of the Lyman series is displaced by
$\mathrm{\Delta}\lambda = 0.20\ nm$ when their radiation is observed at
an angle $\theta = 45{^\circ}$ to their motion direction.

Answer: $7.0 \times 10^{5}\ m/s$

Solution:
Since
$\mathrm{\Delta}\lambda( = 0.20\ nm) \ll \lambda( = 121\ nm)$ of the
first Lyman line of H atom, we need not worry about $v^{2}/c^{2}$
effects. Then
$$\omega^{'} = \frac{\omega}{1 - \beta\cos\theta},\beta = \frac{v}{c}$$
Hence
$1 - \beta\cos\theta = \frac{\omega}{\omega^{'}} = \frac{\lambda^{'}}{\lambda}$
Or
$\beta\cos\theta = 1 - \frac{\lambda^{'}}{\lambda} = \frac{\mathrm{\Delta}\lambda}{\lambda}$
But $\omega = \frac{3}{4}R$ so
$\lambda = \frac{2\pi c}{\omega} = \frac{8\pi c}{3R}$
Hence $v = c\beta = \frac{3R\mathrm{\Delta}\lambda}{8\pi\cos\theta}$
Substitution gives $\left( \cos\theta = \frac{1}{\sqrt{2}} \right)$
$$v = 7.0 \times 10^{5}\ m/s$$

Physics


Question:
A particle moves along an are of a circle of radius $R$
according to the law $l = a\sin{\omega t}$, where $l$ is the
displacement from the initial position measured along the arc, and $a$
and $\omega$ are constants. Assuming $R = 1.00\ m,a = 0.80\ m,$ and
$\omega = 2.00\ rad/s,$ find:
\(a\) the magnitude of the total acceleration of the particle at the
points $l = 0$ and $l = \pm a;$
\(b\) the minimum value of the total acceleration $w_{mn}$ and the
corresponding displacement $l_{m}$.

Answer: $\left| w_{t} \right| = a\omega^{2}$

Solution:
From the equation
$$l = a\sin\omega t$$
$$\frac{dl}{dt} = v = a\omega\cos\omega t$$
So,
$$w_{t} = \frac{dv}{dt} = - a\omega^{2}\sin\omega t\quad(1)$$
and
$$w_{n} = \frac{v^{2}}{R} = \frac{a^{2}\omega^{2}\cos^{2}\omega t}{R}\quad(2)$$
\(a\) At the point $l = 0$, $\sin\omega t = 0$ and
$\cos\omega t = \pm 1$ so, $\omega t = 0,\pi$ etc.
Hence
$$w = w_{n} = \frac{a^{2}\omega^{2}}{R}$$
Similarly at $l = \pm a$, $\sin\omega t = \pm 1$ and
$\cos\omega t = 0$, so, $w_{n} = 0$
Hence
$$w = \left| w_{t} \right| = a\omega^{2}$$

Physics


Question:
Calculate the magnetic moment of a thin wire with a
current $I = 0.8\ A,$ wound tightly on half a tore Fig. The diameter of
the cross-section of the tore is equal to $d = 5.0\ cm$, the number of
turns is $N = 500.$
[IMAGE]

Answer: $\frac{1}{2}Nd^{2}I$

Solution:
Take an element of length $rd\ \theta$ containing
$\frac{N}{\pi r}.rd\theta$ turns. Its magnetic moment is
$$\frac{N}{\pi}d\theta.\frac{\pi}{4}d^{2}I$$
Normal to the plane of cross section. We resolve it along OA and OB.
The moment along $OA$ integrates to
$$\int_{0}^{\pi}{\frac{N}{4}d^{2}Id\theta}\cos\theta = 0$$
while that along $OB$ gives
$$p_{m} = \int_{0}^{\pi}\frac{Nd^{2}I}{4}\sin\theta d\theta = \frac{1}{2}Nd^{2}I$$
[IMAGE]

Physics


Question:
A parallel beam of light falls from vacuum on a surface
enclosing a medium with refractive index $n$ Fig. Find the shape of that
surface, $x(r)$, if the beam is brought into focus at the point $F$ at a
distance $f$ from the crest $O$. What is the maximum radius of a beam
that can still be focussed ?
[IMAGE]

Answer: $r_{\max} = f\sqrt{(n - 1)/(n + 1)}$

Solution:
All rays focusing at a point must have traversed the same
optical path. Thus
$x + n\sqrt{r^{2} + (f - x)^{2}} = nf$ or
$(n - f - x)^{2} = n^{2}r^{2} + n^{2}(f - x)^{2}$
Or,
$n^{2}r^{2} = (nf - x)^{2} - \left\lbrack n(f - x) \right\rbrack^{2} = (nf - x + nf - nx)(nf - x - nf + nx)$
$= x(n - 1)\left( 2nf - (n + 1)x \right)$
$= 2n(n - 1)fx - (n + 1)(n - 1)x^{2}$
Thus, $(n + 1)(n - 1)x^{2} - 2n(n - 1)fx + n^{2}r^{2} = 0$
So,
$x = \frac{n(n - 1)f \pm \sqrt{n^{2}(n - 1)^{2}f^{2} - n^{2}r^{2}(n + 1)(n - 1)}}{(n + 1)(n - 1)}$
$= \frac{nf}{n + 1}\left\lbrack 1 \pm \sqrt{1 - \frac{n + 1}{n - 1}\frac{r^{2}\ }{f^{2}}} \right\rbrack$
Ray must move forward so $x < f,$ for + sign $x > f$ for small $r$, so
--sign.
(Also $x \rightarrow 0$ as $r \rightarrow 0$)
($x > f$ means ray turning back in the direction of incidence.
[IMAGE] Hence
$x = \frac{nf}{n + 1}\left\lbrack 1 - \sqrt{1 - \frac{n + 1}{n - 1}\frac{r^{2}}{f^{2}}} \right\rbrack$
For the maximum value of $r$,
$$\sqrt{1 - \frac{n + 1}{n - 1}\frac{r^{2}}{f^{2}}} = 0$$
because the expression under the radical sign must be non-negative,
which gives the maximum value of $r$.
Hence form Eqn.(A), $r_{\max} = f\sqrt{(n - 1)/(n + 1)}$

Physics


Question:
Three identical discs $A,B,$ and $C$ Fig. rest on a
smooth horizontal plane. The disc $A$ is set in motion with velocity $v$
after which it experiences an elastic collision simultaneously with the
discs B and C. The distance between the centres of the latter discs
prior to the collision is $\eta$ times greater than the diameter of each
disc. Find the velocity of the disc A after the collision. At what value
of $\eta$ will the disc $A$ recoil after the collision; stop; move on?
[IMAGE]

Answer: $\eta = \sqrt{2}$

Solution:
From the symmetry of the problem, the velocity of the disc
$A$ will be directed either in the initial direction or opposite to it
just after the impact. Let the velocity of the disc $A$ after the
collision be $v^{'}$ and be directed towards right after the collision.
It is also clear from the symmetry of problem that the discs $B$ and $C$
have equal speed (say $v"$) in the directions, shown. From the condition
of the problem.
$\cos{\theta = \frac{\eta\frac{d}{2}}{d}} = \frac{\eta}{2}$ so,
$\sin\theta = \sqrt{4 - \eta^{2}}/2$ (1)
For the three discs, system, from the conservation of linear momentum
in the symmetry direction (towards right)
$mv = 2mv"\ sin\theta\ + mv'$ or, $v = 2v"\ sin\ \theta + v'$ (2)
From the definition of the coefficient of restitution, we have for the
discs $A$ and $B$ (or $C$)
$$e = \frac{v" - v'\sin\theta}{v\sin{\theta - 0}}$$
But $e = 1$, for perfectly elastic collision,
So, $v\sin\theta = v" - v'\ sin\ \theta$ (3)
[IMAGE] From (2) and (3),
$$v^{'} = \frac{v\left( 1 - 2\sin^{2}\theta \right)}{\left( 1 + 2\sin^{2}\theta \right)}$$
$= \frac{v\left( \eta^{2} - 2 \right)}{6 - \eta^{2}}$ {using (1)}
Hence, we have,
$$v^{'} = \frac{v\left( \eta^{2} - 2 \right)}{6 - \eta^{2}}$$
Therefore, the disc $A$ will recoil if $\eta < \sqrt{2}$ and stop if
$\eta = \sqrt{2}$.
Note: One can write the equations of momentum conservation along
the direction perpendicular to the initial direction of disc $A$ and
the conservation of kinetic energy instead of the equation of
restitution.

Physics


Question:
In the arrangement shown in Fig. the masses m of the
bar and M of the wedge, as well as the wedge angle α, are known.
[IMAGE] The masses of the pulley and the thread are negligible. The friction
is absent. Find the acceleration of the wedge M.

Answer: $\frac{mg\sin\alpha}{M + 2m\left( 1 - \cos\alpha \right)}$

Solution:
To analyse the kinematic relations between the bodies,
sketch the force diagram of each body as shown in the figure.
On the basis of force diagram, it is obvious that the wedge M will
move towards right and the block will move down along the wedge. As
the length of the thread is constant, the distance travelled by the
block on the wedge must be equal to the distance travelled by the
wedge on the floor. Hence $ds_{mM} = ds_{M}$. As
${\overrightarrow{v}}_{mM}$ and ${\overrightarrow{v}}_{M}$ do not
change their directions and acceleration that's why
${\overrightarrow{w}}_{mM} \uparrow \uparrow {\overrightarrow{v}}_{mM}$
and
${\overrightarrow{w}}_{M} \uparrow \uparrow {\overrightarrow{v}}_{M}$
and $w_{mM} = w_{M} = w$ (say) and accordingly the diagram of
kinematical dependence is shown in figure.
[IMAGE] As
${\overrightarrow{w}}_{m} = {\overrightarrow{w}}_{mM} + {\overrightarrow{w}}_{M}$,
so from triangle law of vector addition.
$w_{m} = \sqrt{w_{M}^{2} + w_{mM}^{2} - 2w_{mM}w_{M}\cos\alpha} = w\sqrt{2\left( 1 - \cos\alpha \right)}$
(1)
From $F_{x} = mw_{x},$ (for the wedge),
$T = T\cos\alpha + N\sin\alpha = Mw$ (2)
For the bar m let us fix $(x - y)$ coordinate system in the frame of
ground Newton's law in projection form along $x$ and $y$ axes Fig.
gives
$$mg\sin\alpha - T = mw_{m(x)} = m\left\lbrack w_{mM(x)} + w_{M(x)} \right\rbrack$$
$$= m\left\lbrack w_{mM} + w_{M}\cos(\pi - \alpha) \right\rbrack = mw(1 - \cos\alpha)$$
$$mg\cos{\alpha - N} = mw_{m(y)} = m\left\lbrack w_{mM(y)} + w_{M(y)} \right\rbrack = m\left\lbrack 0 + w\sin\alpha \right\rbrack$$
Solving the above Eqs. simultaneously, we get
$$w = \frac{mg\sin\alpha}{M + 2m\left( 1 - \cos\alpha \right)}$$
Note: We can study the motion of the block $m$ in the frame of
wedge also, alternately we may solve this problem using conservation
of mechanical energy.

Physics


Question:
A vertical shaft of light from a projector forms a
light spot $S = 100\ cm^{2}$ in area on the ceiling of a round room of
radius $R = 2.0\ m$. The illuminance of the spot is equal to
$E = 1000\ lx$. The reflection coefficient of the ceiling is equal to
$\rho = 0.80$. Find the maximum illuminance of the wall produced by the
light reflected from the ceiling. The reflection is assumed to obey
Lambert's law.

Answer: $0.21\ lux$

Solution:
The illuminated area acts as a Lambert source of
luminosity$M = \pi L$ where
$MS = \rho ES =$ total reflected light
Thus, the luminance
$$L = \frac{\rho E}{\pi}$$
[IMAGE] The equivalent luminous intensity in the direction making an angle
$\theta$ from the vertical is
$$LS\cos\theta = \frac{\rho ES}{\pi}\cos\theta$$
and the illuminance at the point P is
$$\frac{\rho ES}{\pi}\cos\theta\sin\theta/R^{2}{cosec}^{2}\theta = \frac{\rho ES}{\pi R^{2}}\cos\theta\sin^{3}\theta$$
This is maximum when
$$\frac{d}{d\theta}\left( \cos\theta\sin^{3}\theta \right) = - \sin^{4}\theta = 3\sin^{2}\theta\cos^{2}\theta = 0$$
Or $\tan^{2}\theta = 3 \Rightarrow \tan\theta = \sqrt{3}$
Then the maximum illuminance is
$$\frac{3\sqrt{3}}{16\pi}\frac{\rho ES}{R^{2}}$$
This illuminance is obtained at a distance $R\cot\theta = R/\sqrt{3}$
from the ceiling. Substitution gives the value
$$0.21\ lux$$

Physics


Question:
A circuit consists of a capacitor with capacitance $C$
and a coil of inductance $L$ connected in series, as well as a switch
and a resistance equal to the critical value for this circuit. With the
switch disconnected, the capacitor was charged to a voltage $V_{0}$, and
at the moment $t = 0$ the switch was closed. Find the current $I$ in the
circuit as a function of time $t$. What is $I_{\max}$ equal to ?

Answer: $\frac{V_{0}}{e}\sqrt{\frac{C}{L}}$

Solution:
$o = \frac{q}{C} + L\frac{dI}{dt} + RI,I = + \frac{dq}{dt}$
[IMAGE] For the critical case $R = 2\sqrt{\frac{L}{C}}$
Thus $LC\ddot{q} + 2\sqrt{LC}\dot{q} + q = 0$
Look for a solution with $q\alpha e^{\alpha t}$
$$\alpha = - \frac{1}{\sqrt{LC}}$$
An independent solution is $te^{\alpha t}$. Thus
$$q = (A + Bt)e^{- t/\sqrt{LC}},$$
At $t = 0\ q = CV_{0}$ thus $A = CV_{0}$
Also at $t = 0\ \dot{q} = I = 0$
$$0 = B - A\frac{1}{\sqrt{l\ C}} \Rightarrow B = V_{0}\sqrt{\frac{C}{L}}$$
Thus finally
$I = \frac{dq}{dt} = V_{0}\sqrt{\frac{C}{L}}e^{- t/\sqrt{LC}}$
$$- \frac{1}{\sqrt{LC}}\left( CV_{0} + V_{0}\sqrt{\frac{C}{L}}t \right)e^{- t/\sqrt{LC}}$$
$$= - \frac{V_{0}}{L}t\ e^{- t/\sqrt{LC}}$$
The current has been defined to increase the charge. Hence the minus
sign.
The current is maximum when
$$\frac{dI}{dt} = - \frac{V_{0}}{e}e^{- \frac{t}{\sqrt{LC}}}\left( 1 - \frac{t}{\sqrt{LC}} \right) = 0$$
This gives $t = \sqrt{LC}$ and the magnitude of the maximum current is
$$\left| I_{\max} \right| = \frac{V_{0}}{e}\sqrt{\frac{C}{L}}$$

Physics


Question:
A current $I$ flows in a long thin walled cylinder of
radius $R$. What pressure do the walls of the cylinder experience?

Answer: $\frac{\mu_{0}I^{2}}{8\pi^{2}R^{2}}$

Solution:
Let $t =$ thickness of the wall of the cylinder. Then,
$J = I/2\pi Rt$ along z axis. The magnetic field due to this at a
distance $r\left( R - \frac{t}{2} < r < R + \frac{t}{2} \right)$, is
given by,
[IMAGE] $$B_{\varphi}(2\pi r) = \mu_{0}\frac{I}{2\pi Rt}\pi\left\{ r^{2} - \left( R - \frac{t}{2} \right)^{2} \right\}$$
Or,
$B_{\varphi} = \frac{\mu_{0}I}{4\pi Rrt}\left\{ r^{2} - (R - t/2)^{2} \right\}$
Now,
$\overrightarrow{F} = \int_{}^{}{\overrightarrow{J} \times \overrightarrow{B}dV}$
and
$p = \frac{F_{r}}{2\pi RL} = \frac{1}{2\pi RL}\int_{R - \frac{t}{2}}^{R + \frac{t}{2}}\frac{\mu_{0}I^{2}}{8\pi^{2}R^{2}t^{2}r}\left\{ r^{2} - \left( R - \frac{t}{2} \right)^{2} \right\} \times 2\pi rLdr$
$$= \frac{\mu_{0}I^{2}}{8\pi^{2}R^{3}t^{2}}\int_{R - \frac{t}{2}}^{R + \frac{t}{2}}{\left\{ r^{2} - \left( R - \frac{t}{2} \right)^{2} \right\} dr} = \frac{\mu_{0}I^{2}}{8\pi^{2}R^{3}t^{2}}\left\lbrack \frac{\left( R + \frac{t}{2} \right)^{3} - \left( R - \frac{t}{2} \right)^{3}}{3} - \left( R - \frac{t}{2} \right)^{2}t \right\rbrack$$
$$= \frac{\mu_{0}I^{2}}{8\pi^{2}R^{3}t}\left\lbrack Rt + 0\left( t^{2} \right) \right\rbrack = \frac{\mu_{0}I^{2}}{8\pi^{2}R^{2}}$$

Physics


Question:
A square wire frame with side a and straight conductor
carrying a constant current $I$ are located in the same plane Fig.
[IMAGE] The inductance and the resistance of the frame are equal to L and R
respectively. The frame was turned through $180{^\circ}$ about the axis
$OO'$ separated from the current-carrying conductor by a distance $b$.
Find the electric charge having flown through the frame.

Answer: $\frac{\mu_{0}ai}{2\pi R}\ln\frac{b + a}{b - a}$

Solution:
According to Ohm's law and Faraday's law of induction, the
current $i_{0}$ appearing in the frame, during its rotation, is
determined by the formula,
$$i_{0} = - \frac{d\Phi}{dt} = - \frac{Ldi_{0}}{dt}$$
Hence, the required amount of electricity (charge) is,
$$q = \int_{}^{}{i_{0}dt} = - \frac{1}{R}\int_{}^{}\left( d\Phi + Ldi_{0} \right) = - \frac{1}{R}\left( \mathrm{\Delta}\Phi + L\mathrm{\Delta}i_{0} \right)$$
[IMAGE] Since the frame has been stopped after rotation, the current in it
vanishes, and hence $\mathrm{\Delta}i_{0} = 0$. It remains for us to
find the increment of the flux $\mathrm{\Delta}\Phi$ through the frame
$\left( \mathrm{\Delta}\Phi = \Phi_{2} - \Phi_{1} \right)$.
Let us choose the normal $\overrightarrow{n}$ to the plane of the
frame, for instance, so that in the final position,
$\overrightarrow{n}$ is directed behind the plane of the figure (along
$\overrightarrow{B}$).
Then it can be easily seen that in the final position, $\Phi_{2} > 0$,
while in the initial position, $\Phi_{1} < 0$ (the normal is opposite
to $\overrightarrow{B}$), and $\mathrm{\Delta}\Phi$ turns out to be
simply equal to the fulx through the surface bounded by the final and
initial positions of the frame :
$$\mathrm{\Delta}\Phi = \Phi_{2} + \left| \Phi_{1} \right| = \int_{b - a}^{b + a}{Ba\ dr},$$
where $B$ is a function of r, whose form can be easily found with the
help of the theorem of circulation. Finally omitting the minus sign,
we obtain,
$$q = \frac{\mathrm{\Delta}\Phi}{R} = \frac{\mu_{0}ai}{2\pi R}\ln\frac{b + a}{b - a}$$

Physics


Question:
A horizontal smooth rod $AB$ rotates with a constant
angular velocity $\omega = 2.00\ rad/s$ about a vertical axis passing
through its end $A$. A feely sliding sleeve of mass m = 0.50 kg moves
along the rod from the point A with the initial velocity
$v_{0} = 1.00\ m/s,$. Find the Coriolis force acting on the sleeve (in
the reference frame fixed to the rotating rod) at the moment when the
sleeve is located at the distance $r = 50\ cm$ from the rotation axis.

Answer: 2.83 N

Solution:
The sleeve is free to slide along the rod $AB$. Thus only
the centrifugal force acts on it.
The equation is,
$m\dot{v} = m\omega^{2}r$ where $v = \frac{dr}{dt}$
But
$\dot{v} = v\frac{dv}{dt} = \frac{d}{dr}\left( \frac{1}{2}v^{2} \right)$
So, $\frac{1}{2}v^{2} = \frac{1}{2}\omega^{2}r^{2} +$ constant
Or, $v^{2} = v_{0}^{2} + \omega^{2}r^{2}$
$v_{0}$ being the initial velocity when $r = 0$. The Coriolis force is
then,
$$2m\omega\sqrt{v_{0}^{2} + \omega^{2}r^{2}} = 2m\omega^{2}r\sqrt{1 + v_{0}^{2}/\omega^{2}r^{2}}$$
= 2.83 N on putting the values.

Physics


Question:
A horizontal plane supports stationary vertical
cylinder of radius $R$ and a disc $A$ attached to the cylinder by a
horizontal thread $AB$ of length $l_{0}$ (Fig. top view). An initial
velocity $v_{0}$ is imparted to the disc as shown in the figure. How
long will it move along the plane until it strikes against the cylinder?
The friction is assumed to be absent.
[IMAGE]

Answer: $\frac{l_{0}^{2}}{2Rv_{0}}$

Solution:
Since the tension is always perpendicular to the velocity
vector, the work done by the tension force will be zero. Hence,
according to the work energy theorem, the kinetic energy or velocity of
the disc will remain constant during it's motion. Hence, the sought time
$t = \frac{s}{v_{0}}$, where $s$ is the total distance traversed by the
small disc during it's motion.
Now, at an arbitrary position Fig.
[IMAGE] $$ds = \left( l_{0} - R\theta \right)d\theta,$$
So, $s = \int_{0}^{l/R}\left( l_{0} - R\theta \right)d\theta,$
Or,
$s = \frac{l_{0}^{2}}{R} - \frac{Rl_{0}^{2}}{2R^{2}} = \frac{l_{0}^{2}}{2R}$
Hence, the required time, $t = \frac{l_{0}^{2}}{2Rv_{0}}$
It should be clearly understood that the only uncompensated force
acting on the disc A in this case is the tension $T$, of the thread.
It is easy to see that there is no point here, relative to which the
moment of force $T$ is invariable in the process of motion. Hence
conservation of angular momentum is not applicable here.

Physics


Question:
There are two cavities Fig. with small holes of equal
diameters $d = 1.0\ cm$ and perfectly reflecting outer surfaces. The
distance between the holes is $l = 10\ cm$. A constant temperature
$T_{1} = 1700\ K$ is maintained in cavity 1. Calculate the steady-state
temperature inside cavity 2.
Instruction: Take into account that a black body radiation obeys
the cosine emission law.
[IMAGE]

Answer: $380\ K$

Solution:
Taking account of cosine low of emissions we write for the
energy radiated per second by the hole in cavity \# 1 as
$$dI(\Omega) = A\cos\theta d\Omega$$
where $A$ is an constant, $d\Omega$ is an element of solid angle
around some direction defined by the symbol $\Omega$. Integrating over
the whole forward hemisphere we get
$$I = A\int_{0}^{\pi/2}{\cos\theta 2\pi\sin\theta d\theta} = \pi A$$
[IMAGE] We find $A$ by equating this to the quantity
$\sigma T_{1}^{4} \bullet \frac{\pi d^{2}}{4}\sigma$ is
Stefan-Boltaman constant and $d$ is the diameter of the hole.
Then $A = \frac{1}{4}\sigma d^{2}T_{1}^{4}$
Now energy reaching 2 from 1 is
$\left( \cos{\theta \approx 1} \right)$
$$\frac{1}{4}\sigma d^{2}T_{1}^{4} \bullet \mathrm{\Delta}\Omega$$
where
$\mathrm{\Delta}\Omega = \frac{\left( \pi d^{2}/4 \right)}{l^{2}}$ is
the solid angle subtended by the hole of 2 at 1. {We are assuming
$d \ll l$ so $\mathrm{\Delta}\Omega =$ area of hole / (distance)^2^}
This must equal $\sigma T_{2}^{4}\pi d^{2}/4$
which is the energy emitted by 2. Thus equating
$$\frac{1}{4}\sigma d^{2}T_{1}^{4}\frac{\pi d^{2}}{4l^{2}} = \sigma T_{2}^{4}\frac{\pi d^{2}}{4}$$
Or $T_{2} = T_{1}\sqrt{\frac{d}{2l}}$
Substituting we get $T_{2} = 0.380\ kK = 380\ K$

Physics


Question:
A light ray composed of two monochromatic components
passes through a trihedral prism with refracting angle
$\theta = 60{^\circ}$. Find the angle $\mathrm{\Delta}\alpha$ between
the components of the ray after its passage through the prism if their
respective indices of refraction are equal to 1.515 and 1.520. The prism
is oriented to provide the least deflection angle.

Answer: $0.44$

Solution:
From the Caubhy's formula, and also experimentally the
R.I. of a medium depends upon the wavelength of the mochromatic ray i.e.
$n = f(\lambda)$. In the case of least deviation of a monochromatic ray
the passage a prism, we have :
$n\sin{\frac{\theta}{2} = \sin\frac{\alpha + \theta}{2}}$ (1)
The above equation tales us that we have $n = n(\alpha)$, so we may
write
$\mathrm{\Delta}n = \frac{dn}{d\alpha}\mathrm{\Delta}\alpha$ (2)
From Eqn. (1)
$$dn\sin\frac{\theta}{2} = \frac{1}{2}\cos\frac{\alpha + \theta}{2}d\alpha$$
Or,
$\frac{dn}{d\alpha} = \frac{\cos\frac{\alpha + \theta}{2}}{2\sin\frac{\theta}{2}}$
(3)
[IMAGE] From Eqns (2) and (3)
$$\mathrm{\Delta}n = \frac{\cos\frac{\alpha + \theta}{2}}{2\sin\frac{\theta}{2}}\mathrm{\Delta}\alpha$$
Or,
$\mathrm{\Delta}n = \frac{\sqrt{1 - \sin^{2}\left( \frac{\alpha + \theta}{2} \right)}}{2\sin\frac{\theta}{2}}\mathrm{\Delta}\alpha = \frac{\sqrt{1 - n^{2}\sin^{2}\frac{\theta}{2}}}{2\sin\frac{\theta}{2}}\mathrm{\Delta}\alpha$
(Using Eqn. 1.)
Thus
$\mathrm{\Delta}\alpha = \frac{2\sin\frac{\theta}{2}}{\sqrt{1 - n^{2}}\sin^{2}\frac{\theta}{2}}\mathrm{\Delta}n = 0.44$

Physics


Question:
A rotating disc Fig. moves in the positive direction of
the $x$ axis. Find the equation $y(x)$ describing the position of the
instantaneous axis of rotation, if at the initial moment the axis $C$ of
the disc was located at the point O after which it moved
\(a\) with a constant velocity $v$, while the disc started rotating
counterclockwise with a constant angular acceleration $\beta$ (the
initial angular velocity is equal to zero);
\(b\) with a constant acceleration $w$ (and the zero initial
velocity), while the disc rotates counterclockwise with a constant
angular velocity $\omega$.
[IMAGE]

Answer:
(a) $\frac{v^{2}}{\beta x}$
\(b\) $\frac{\sqrt{2wx}}{\omega}$

Solution:
Rotating disc moves along the $x$-axis, in plane motion in
$x - y$ plane. Plane motion of a solid
can be imagined to be in pure rotation about a point (say $I$) at a
certain instant known as instantaneous centre of rotation. The
instantaneous axis whose positive sense is directed along
$\overrightarrow{\omega}$ of the solid and which passes through the
point $I$, is known as instantaneous axis of rotation.
Therefore the velocity vector of an arbitrary point $(P)$ of the solid
can be represented as:
$${\overrightarrow{v}}_{P} = \overrightarrow{\omega} \times {\overrightarrow{r}}_{PI}\quad(1)$$
On the basis of Eq. (1) for the C. M. $(C)$ of the disc
$${\overrightarrow{v}}_{c} = \overrightarrow{\omega} \times {\overrightarrow{r}}_{cI}\quad(2)$$
According to the problem
${\overrightarrow{v}}_{c} \uparrow \uparrow \overrightarrow{i}$ and
$\overrightarrow{\omega} \uparrow \uparrow \overrightarrow{k}$
i.e. $\overrightarrow{\omega}\bot x - y$ plane, so to satisfy the Eqn.
(2)
${\overrightarrow{r}}_{cI}$ is directed along
$\left( - \overrightarrow{j} \right)$. Hence point $I$ is at a
distance $r_{CI} = y$, above the centre of the disc
along $y$-axis. Using all these facts in Eq. (2), we get
$$v_{c} = \omega y\quad\text{or}\quad y = \frac{v_{c}}{\omega}\quad(3)$$
[IMAGE] Figure 1.51: Insert this diagram from the shared reference image ---
instantaneous centre of rotation for the rolling disc.
(a) From the angular kinematical equation
$$\omega_{z} = \omega_{0z} + \beta_{z}t\quad(4)$$
$$\omega = \beta t.$$
On the other hand $x = vt$, (where $x$ is the $x$ coordinate of the
C.M.)
or,
$$t = \frac{x}{v}\quad(5)$$
From Eqs. (4) and (5),
$$\omega = \frac{\beta x}{v}$$
Using this value of $\omega$ in Eq. (3) we get
$$y = \frac{v_{c}}{\omega} = \frac{v}{\beta x/v} = \frac{v^{2}}{\beta x}$$
(hyperbola)
(b) As centre $C$ moves with constant acceleration $w$, with zero
initial velocity
So,
$$x = \frac{1}{2}wt^{2}\quad\text{and}\quad v_{c} = wt$$
Therefore,
$$v_{c} = w\sqrt{\frac{2x}{w}} = \sqrt{2xw}$$
Hence
$$y = \frac{v_{c}}{\omega} = \frac{\sqrt{2wx}}{\omega}$$
(parabola)

Physics


Question:
A parallel-plate capacitor is located horizontally so
that one of its plates is submerged into liquid while the other is over
its surface Fig. The permittivity of the liquid is equal to
$\varepsilon$, its density is equal to $\rho$. To what height will the
level of the liquid in the capacitor rise after its plates get a charge
of surface density $\sigma$?
[IMAGE]

Answer: $h = \frac{(\varepsilon - 1)\sigma^{2}}{2\varepsilon_{0}\varepsilon\rho g}$

Solution:
One way of doing this problem will be exactly as in the
previous case so let us try an alternative method based on energy.
Suppose the liquid rises by a distance $h$. Then let us calculate the
extra energy of the liquid as a sum of polarization energy and the
ordinary gravitational energy. The latter is
$$\frac{1}{2}h \bullet \rho g \bullet Sh = \frac{1}{2}\rho gSh^{2}$$
If $\sigma$ is the free charge surface density on the plate, the bound
charge density is, from the previous problem,
$$\sigma^{'} = \frac{\varepsilon - 1}{\varepsilon}\sigma$$
This is also the volume density of induced dipole moment i.e.
Polarization. Then the energy is, as before
$$- \frac{1}{2} \bullet \sigma^{'}E_{0} = \frac{- 1}{2} \bullet \sigma^{'}\frac{\sigma}{\varepsilon_{0}} = \frac{- (\varepsilon - 1)\sigma^{2}}{2\varepsilon_{0}\varepsilon}$$
and the total polarization energy is
$$- S(a + h)\frac{(\varepsilon - 1)\sigma^{2}}{2\varepsilon_{0}\varepsilon}$$
[IMAGE] Then, total energy is
$$U(h) = - S(a + h)\frac{(\varepsilon - 1)\sigma^{2}}{2\varepsilon_{0}\varepsilon} + \frac{1}{2}\rho gSh^{2}$$
The actual height to which the liquid rises is determined from the
formula
$$\frac{dU}{dh} = U^{'}(h) = 0$$
This gives
$h = \frac{(\varepsilon - 1)\sigma^{2}}{2\varepsilon_{0}\varepsilon\rho g}$

Physics


Question:
A particle of mass $m$ is located outside a uniform
sphere of mass $M$ at a distance $r$ from its centre. Find:
\(a\) the potential energy of gravitational interaction of the particle
and the sphere;
\(b\) the gravitational force which the sphere exerts on the particle.

Answer:
(a) $- \frac{\gamma Mm}{r}$
\(b\) $- \frac{\gamma mM}{r^{3}}\overrightarrow{r}$

Solution:
Suppose that the sphere has a radius equal to $a$. We may
imagine that the sphere is made up of concentric thin spherical shells
(layers) with radii ranging from 0 to a, and each spherical layer is
made up of elementary bands (rings). Let us first calculate potential
due to an elementary band of spherical layer at the point of location of
the point mass $m$ (say point $P$) Fig. As all the points of the band
are located at the distance $l$ from the point $P$, so,
[IMAGE] $\partial\varphi = - \frac{\gamma\partial M}{l}$ (where mass of the
band) (1)
$$\partial M = \left( \frac{dM}{4\pi a^{2}} \right)\left( 2\pi a\sin\theta \right)(a\ d\theta)$$
$= \left( \frac{dM}{2} \right)\sin\theta d\theta$ (2)
And $l^{2} = a^{2} + r^{2} - 2ar\cos\theta$ (3)
Differentiating Eq. (3), we get
$ldl = ar\sin\theta d\theta$ (4)
Hence using above equations
$\partial\varphi = - \left( \frac{\gamma dM}{2ar} \right)dl$ (5)
Now integrating this Eq. over the whole spherical layer
$$d\varphi = \int_{}^{}{\partial\varphi} = - \frac{\gamma dM}{2ar}\int_{r - a}^{r + a}\ $$
So $d\varphi = - \frac{\gamma dM}{r}$ (6)
Equation (6) demonstrates that the potential produced by a thin
uniform spherical layer outside the layer is such as if the whole mass
of the layer were concentrated at it's centre;
Hence the potential due to the sphere at point P;
$\varphi = \int_{}^{}{d\varphi} = - \frac{\gamma}{r}\int_{}^{}{dM} = - \frac{\gamma M}{r}$
(7)
This expression is similar to that of Eq. (6)
Hence the sought potential energy of gravitational interaction of the
particle $m$ and the sphere,
$$U = m\varphi = - \frac{\gamma Mm}{r}$$
\(b\) Using the Eq., $G_{r} = - \frac{\partial\varphi}{\partial r}$
$G_{r} = - \frac{\gamma M}{r^{2}}$ (using Eq. 7)
So, $\overrightarrow{G} = \frac{\gamma M}{r^{3}}\overrightarrow{r}$
and
$\overrightarrow{F} = m\overrightarrow{G} = - \frac{\gamma mM}{r^{3}}\overrightarrow{r}$
(8)

Physics


Question:
Two bars connected by a weightless spring of stiffness
$x$ and length (in the non-deformed state) $l_{0}$ rest on a horizontal
plane. A constant horizontal force $F$ starts acting on one of the bars
as shown in Fig. Find the maximum and minimum distances between the bars
during the subsequent motion of the system, if the masses of the bars
are:
\(a\) equal;'
\(b\) equal to $m_{1}$ and $m_{2}$, and the force $F$ is applied to
the bar of mass $m_{2}$.
[IMAGE]

Answer: $l_{0} + \frac{2m_{1}F}{k\left( m_{1} + m_{2} \right)}$

Solution:
Let us consider both blocks and spring as the physical
system. The centre of mass of the system moves with acceleration
$a = \frac{F}{m_{1} + m_{2}}$ towards right. Let us work in the frame of
centre of mass. As this frame is a non-inertial frame (accelerated with
respect to the ground) we have to apply a pseudo force $m_{1}a$ towards
left on the block $m_{1}$ and $m_{2}$ a towards left on the block
$m_{2}$.
As the center of mass is at rest in this frame, the blocks move in
opposite directions and come to instantaneous rest at some instant.
The elongation of the spring will be maximum or minimum at this
instant. Assume that the block $m_{1}$ is displaced by the distance
$x_{1}$ and the block $m_{2}$ through a distance $x_{2}$ from the
initial positions.
[IMAGE] From the energy equation in the frame of C.M.
$$\mathrm{\Delta}\widetilde{T} + U = A_{ext},$$
(where $A_{ext}$ also includes the work done by the pseudo forces)
Here,
$\mathrm{\Delta}\widetilde{T} = 0,U = \frac{1}{2}k\left( x_{1} + x_{2} \right)^{2}$
and
$$W_{ext} = \left( \frac{F - m_{2}F}{m_{1} + m_{2}} \right)x_{2} + \frac{m_{1}F}{m_{1} + m_{2}}x_{1} = \frac{m_{1}F\left( x_{1} + x_{2} \right)}{m_{1} + m_{2}},$$
Or,
$\frac{1}{2}k\left( x_{1} + x_{2} \right)^{2} = \frac{m_{1}\left( x_{1} + x_{2} \right)F}{m_{1} + m_{2}}$
So, $x_{1} + x_{2} = 0$ or,
$x_{1} + x_{2} = \frac{2m_{1}F}{k\left( m_{1} + m_{2} \right)}$
Hence the maximum separation between the blocks equals:
$l_{0} + \frac{2m_{1}F}{k\left( m_{1} + m_{2} \right)}$
Obviously the minimum sepation corresponds to zero elongation and is
equal to $l_{0}$

Physics


Question:
A horizontally oriented tube $AB$ of length $l$ rotates
with a constant angular velocity $\omega$ about a stationary vertical
axis $OO'$ passing through the end $A$ Fig. The tube is filled with an
ideal fluid. The end $A$ of the tube is open, the closed end $B$ has a
very small orifice. Find the velocity of the fluid relative to the tube
as a function of the column "height" $h$.
[IMAGE]

Answer: $v = \omega h\sqrt{\frac{2l}{h} - 1}$

Solution:
In a rotating frame (with constant angular velocity) the
Eulerian equation is
$$- \overrightarrow{\nabla}p + p\overrightarrow{g} + 2\rho\left( \overrightarrow{v} \times \overrightarrow{\omega} \right) + \rho\omega^{2}\overrightarrow{r} = \rho\frac{d\overrightarrow{v}}{dt}$$
In the frame of rotating tube the liquid in the "column" is
practically static because the orifice is sufficiently small. Thus the
Eulerian Eq. in projection form along $\overrightarrow{r}$ (which is
the position vector of an arbitrary liquid element of length $dr$
relative to the rotation axis) reduces to
[IMAGE] $$\frac{- dp}{dr} + \rho\omega^{2}r = 0$$
Or, $dp = \rho\omega^{2}rdr$
So, $\int_{p_{0}}^{p}{dp} = \rho\omega^{2}\int_{(l - h)}^{r}{rdr}$
Thus
$p(r) = p_{0} + \frac{\rho\omega^{2}}{2}\left\lbrack r^{2} - (l - h)^{2} \right\rbrack$
(1)
Hence the pressure at the end $B$ just before the orifice i.e.
$p(l) = p_{0} + \frac{\rho\omega^{2}}{2}\left( 2lh - h^{2} \right)$
(2)
Then applying Bernoull's theorem at the orifice for the points just
inside and outside of the end $B$
$p_{0} + \frac{1}{2}\rho\omega^{2}\left( 2lh - h^{2} \right) = p_{0} + \frac{1}{2}\rho v^{2}$
(where $v$ is the sought velocity)
So, $v = \omega h\sqrt{\frac{2l}{h} - 1}$

Physics


Question:
In an oscillating circuit shown in Fig. the coil
inductance is equal to $L = 2.5\ mH$ and the capacitor have capacitances
$C_{1} = 2.0\ \mu F$ and $C_{2} = 3.0\ \mu F$. The capacitor were
charged to a voltage $V = 180\ V$, and then the switch $Sw$ was closed.
Find :
\(a\) the natural oscillation frequency;
\(b\) the peak value of the current flowing through the coil.
[IMAGE]

Answer: $0.7\ ms,\ 8.05\ A$

Solution:
The equation of the $L - C$ circuit are
$$L\frac{d}{dt}\left( I_{1} + I \right) = \frac{C_{1}V - \int_{}^{}{I_{1}dt}}{C_{1}} = \frac{C_{2}V - \int_{}^{}I_{2}dt}{C_{2}}$$
Differentiating again
$L\left( I_{1} + I_{2} \right) = - \frac{1}{C_{1}}I_{1} = - \frac{1}{C_{2}}I_{2}$
Then
$I_{1} = \frac{C_{1}}{C_{1} - C_{2}}I,I_{2} = \frac{C_{2}}{C_{1} + C_{2}}I,$
$$I = I_{1} + I_{2}$$
so $L\left( C_{1} + C_{2} \right)I + I = 0$
or $I = I_{0}\sin\left( \omega_{0}t + \alpha \right)$
where $\omega_{0}^{2} = \frac{1}{L\left( C_{1} + C_{2} \right)}$ (Part
a)
(Hence $T = \frac{2\pi}{\omega_{0}} = 0.7\ ms$)
[IMAGE] At $t = 0,I = 0$ so $\alpha = 0$
$$I = I_{0}\sin{\omega_{0}t}$$
The peak value of the current is $I_{0}$ and it is related to the
voltage $V$ by the first equation
$$LI = V - \int_{}^{}Idt/\left( C_{1} + C_{2} \right)$$
Or
$+ L\omega_{0}I_{0}\cos\omega_{0}t = V - \frac{1}{C_{1} + C_{2}}\int_{\omega_{0}}^{t}I_{0}\sin\omega_{0}dt$
(The P.D. across the inductance is $V$ at $t = 0$)
$$= V + \frac{1}{C_{1} + C_{2}} \bullet \frac{I_{0}}{\omega_{0}}\left( \cos\omega_{0}t - 1 \right)$$
Hence
$I_{0} = \left( C_{1} + C_{2} \right)\omega_{0}V = V\sqrt{\frac{C_{1} + C_{2}}{L}} = 8.05\ A$

Physics


Question:
A photon is scattered at an angle
$\theta = 120{^\circ}$ by a stationary free electron. As a result, the
electron acquires a kinetic energy $T = 0.45\ MeV$. Find the energy that
the photon had prior to scattering.

Answer: $0.677\ Mev$

Solution:
From the Compton formula
$$\lambda = \lambda_{0} + \frac{2\pi h}{m\ c}(1 - cos\theta)$$
From conservation of energy
$$\frac{2\pi hc}{\lambda_{0}} = \frac{2\pi hc}{\lambda} + T = \frac{2\pi hc}{\lambda_{0} + \frac{2\pi h}{mc}\left( 1 - \cos\theta \right)} + T$$
Or
$\frac{4\pi h}{mc}\sin^{2}\frac{\theta}{2} = \frac{T}{2\pi hc}\lambda_{0}\left( \lambda_{0} + \frac{4\pi h}{m\ c}\sin^{2}\frac{\theta}{2} \right)$
or introducing $h\omega_{0} = 2\pi hc/\lambda_{0}$
$$\frac{2\sin^{2}{\theta/2}}{mc^{2}} = \frac{T}{h\omega_{0}}\left( \frac{1}{h\omega_{0}} + \frac{2}{mc^{2}}\sin^{2}\frac{\theta}{2} \right)$$
Hence
$\left( \frac{1}{h\omega_{0}} \right)^{2} + 2\frac{1}{h\omega_{0}}\frac{\sin^{2}\frac{\theta}{2}}{mc^{2}} - \frac{2\sin^{2}\frac{\theta}{2}}{mc^{2}T} = 0$
$$\left( \frac{1}{h\omega_{0}} + \frac{\sin^{2}\frac{\theta}{2}}{mc^{2}} \right)^{2} = \frac{2\sin^{2}\frac{\theta}{2}}{mc^{2}T} + \left( \frac{\sin^{2}\frac{\theta}{2}}{mc^{2}\ } \right)^{2}$$
$$\frac{1}{h\omega_{0}} = \frac{\sin^{2}\frac{\theta}{2}}{mc^{2}}\left\lbrack \sqrt{1 + \frac{2mc^{2}}{T\sin^{2}\frac{\theta}{2}}} - 1 \right\rbrack$$
Or
$h\omega_{0} = \frac{mc^{2}/\sin^{2}{\theta/2}}{\sqrt{1 + \frac{2mc^{2}}{T\sin^{2}\frac{\theta}{2}}} - 1}$
$$= \frac{T}{2}\left\lbrack \sqrt{1 + \frac{2mc^{2}}{T\sin^{2}{\theta/2}}} + 1 \right\rbrack$$
Substituting we get $h\omega_{0} = 0.677\ Mev$

Physics


Question:
An ideal gas of molar mass $M$ is contained in a very
tall vertical cylindrical vessel in the uniform gravitational field in
which the free-fall acceleration equals $g$. Assuming the gas
temperature to be the same and equal to $T$, find the height at which
the centre of gravity of the gas is located.

Answer: $h = \frac{RT}{Mg}$

Solution:
As the gravitational field is constant the centre of
gravity and the centre of mass are same.
The location of C.M.
$$h = \frac{\int_{0}^{\infty}hdm}{\int_{0}^{\infty}{dm}} = \frac{\int_{0}^{\infty}{h\rho}dh}{\int_{0}^{\infty}\rho dh}$$
But from Barometric formula and gas law $\rho = \rho_{0}e^{- Mgh/RT}$
So,
$h = \frac{\int_{0}^{\alpha}{h\left( e^{- Mgh/RT} \right)}dh}{\int_{0}^{\alpha}\left( e^{- Mg\ h/RT} \right)dh} = \frac{RT}{Mg}$

Physics


Question:
An object is placed in front of convex surface of a
glass plano-convex lens of thickness $d = 9.0\ cm$. The image of that
object is formed on the plane surface of the lens serving as a screen.
Find:
(a) the transverse magnification if the curvature radius of the lens's
 convex surface is
$$R = 2.5\ cm;$$
(b) the image illuminance if the luminance of the object is
 $L = 7700\ cd/m^{2}$ and the entrance aperture diameter of the lens
 is $D = 5.0\ mm;$ losses of light are negligible.

Answer:
(a) $1 - \frac{d(n - 1)}{nR}$
(b) $42\ 1x$

Solution:
(a) The formation of the image of a source $S$, placed at
a distance $u$ from the pole of the convex surface of plano-convex lens
of thickness $d$ is shown in the figure.
On applying the formula for refraction through spherical surface, we
get
$\frac{n}{s'} - \frac{1}{s} = (n - 1)/R$, (here $n_{2} = n$ and
$n_{1} = 1$)
Or, $\frac{n}{d} - \frac{1}{s} = (n - 1)/R$ or
$\frac{1}{s} = \frac{n}{d} - \frac{(n - 1)}{R}$
Or,
$\frac{s'}{s} = s^{'}\left\{ \frac{n}{d} - \frac{(n - 1)}{R} \right\}$
[IMAGE] But in this case optical path of the light, corresponding to the
distance $v$ in the medium is $v/n$, so the magnification produced
will be,
$$\beta = \frac{s'}{ns} = \frac{s'}{n}\left\{ \frac{n}{d} - \frac{(m - 1)}{R} \right\} = \frac{d}{n}\left\{ \frac{n}{d} - \frac{(n - 1)}{R} \right\} = 1 - \frac{d(n - 1)}{nR}$$
Substituting the values, we get magnification $\beta = - 0.20.$
\(b\) If the transverse area of the object is $A$ (assumed small), the
area of the image is $\beta^{2}A$.
We shall assume that $\frac{\pi D^{2}}{4} > A$. Then light falling on
the lens is : $LA\frac{\pi D^{2}/4}{s^{2}}$
from the definition of luminance (See Eqn. (5.1 c) of the book; here
$\cos\theta \approx 1$ if $D^{2} \ll s^{2}$ and
$d\Omega = \frac{\pi D^{2}/4}{s^{2}}$. Then the illuminance of the
image is
$LA\frac{\pi D^{2}/4}{s^{2}}/\beta^{2}A = Ln^{2}\pi D^{2}/4d^{2}$
Substitution gives 42 1x.

Physics